I'm trying to prove the following:
If $R$ is a Noetherian ring, $M$ a finitely generated module, then $M$ is torsionfree iff the associated primes of $M$ are contained in the associated primes of $R$.
I tried to prove the forward direction, but got confused. My attempt: Let $P$ be an associated prime of $M$. Then there is a non-zero element $x\in M$ such that $P=ann(x)$. Let $p_1,...,p_n$ be generators of $P$. For any $i$, since $p_ix=0$ and $M$ is torsionfree, then $p_i$ must be a zero-divisor in $R$, i.e., there must be $0\neq r_i\in R$ such that $p_ir_i=0$. I claim that $p=ann(r_1\cdots r_n)$. I dont know if this will be the case, but I also dont know if $r_1\cdots r_n=0$ in which case, the my idea wont be helpful. Any ideas are welcomed!